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I love origami, and it recently gave me an idea for a very hard but beautiful puzzle. I'm really curious whether anyone here can solve it.

So here's the puzzle. You are given a large perfectly square piece of paper with no marks on it. With this square, you have to make a square of exactly one-fifth the area of the original square. You are given no tools such as a ruler or scissors, and all you can do is fold the paper. How do you solve this?

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    $\begingroup$ I would suggest "one fifth the size" as "five times smaller" doesn't really make sense. ell.stackexchange.com/a/96873/76722 $\endgroup$ – nasch Aug 17 at 2:48
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    $\begingroup$ @nasch For me 5 times smaller makes sense. The issue in the top answer (not the question) in the linked post is the confusion from having three items being compared ($10, $8, and the new one). For this one it is clear (also from the fact that you are sure that OP means new area equals 0.2 of the old area) since we are comparing only two objects. The convention that I often see is that when we use multiples, we simple multiply/divide by the number, when we use percentage, we do addition/subtraction. So twice larger does not mean 3x the original, but 200% larger is 3x the original. $\endgroup$ – justhalf Aug 17 at 4:33
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    $\begingroup$ @justhalf It doesn't make sense to say "five TIMES smaller" because you need to DIVIDE by five. "Times" means multiplication, not division. If you are going to multiply by something it would be one fifth, so it's one fifth times the size, which is correct but extremely awkward at best. It can never really be "x times smaller" because there is no unit of smallness that can have a greater magnitude. It's a unit of largeness with a smaller magnitude. $\endgroup$ – nasch Aug 17 at 14:03
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    $\begingroup$ @marshalcraft : I'm not a native speaker, and if my original wording was confusing or ambiguous, that wasn't my intent. The puzzle is intended to be a purely imagination puzzle, not a puzzle whose point is a tricky wording. $\endgroup$ – Mitsuko Aug 17 at 18:51
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    $\begingroup$ @nasch you are being pedantic just for the sake of it. "fives times smaller" is neither ambiguous nor grammatically incorrect. It is clear what is meant. Moreover, if you want to get mathematical about it you are also wrong as you have defined neither the statement "five times smaller" or "five times larger". You don't need to "divide anything by five", besides division is defined as multiplication of the inverse, which could mean "smaller". It is perfectly valid to define "$a$ is $n$ times smaller than $b$ if $na=b$". $\endgroup$ – epiliam Aug 19 at 3:13
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The way to do this is:

- Fold the paper in half along both axes. You've now marked the midpoint of all four sides.

- Fold along the knight's-move diagonals, drawn here:

enter image description here

This creates the red square. All five colored regions have the same area, so the red square is 1/5 the size of the square you started with.

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    $\begingroup$ In case anyone else is wondering, it took me a minute to figure this out. You can make those diagonal folds because you have the horizontal and vertical centerlines marked with creases. Fold between (for example) the upper right corner and the bottom center. $\endgroup$ – nasch Aug 17 at 2:52
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    $\begingroup$ @nasch thank you - that's totally an optical illusion that the opposite intersections are not at the same height! $\endgroup$ – rrauenza Aug 17 at 4:26
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    $\begingroup$ Proof: consider the triangle consisting of both yellow tiles + the small teal triangle. Assuming a 1x1 square, the total area is (1/2)(1/2)(1) = 1/4. The small triangles are each 1/4th the area of the larger colored triangles (since they're similar triangles with half the base); thus the large yellow triangle takes up 4/5th of that area, or (4/5)(1/4) = 1/5 total area. So all triangles together = 4/5 area, leaving 1/5 for the red square. $\endgroup$ – BlueRaja - Danny Pflughoeft Aug 17 at 23:43
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    $\begingroup$ @BlueRaja-DannyPflughoeft The proof I had in mind was easier (for me at least): You can reassemble the two shapes of each color into a square the same size as the red. (This is a square because (1) all four angles are right angles; (2) two adjacent sides are the same, because each of the four lines touching the corners are the same by symmetry.) $\endgroup$ – Deusovi Aug 18 at 1:16
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    $\begingroup$ @Chris We can see that e.g the large teal triangle is similar, but double the size of, the small teal triangle, because we know the small triangle's base is half as long. Given that, we can see that the red square runs along exactly half of the large teal triangle's side, because it ends where the small triangle begins. $\endgroup$ – TenMinJoe Aug 18 at 12:09
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Fold the paper horizontally exactly in the middle; fold each of the two $1\times\frac 1 2$ rectangles diagonally such that the two diagonals are parallel. Rotate the paper by a quarter turn and do exactly the same. The four diagonals you have just created enclose a square of area $\frac 1 5$.

We need to show that the distance between two parallel diagonals is $\frac 1 {\sqrt 5}$. This distance equals the height over the diagonal of one of the large triangles we have created. These triangles have area $\frac 1 4$ while the base length i.e. the length of a diagonal is $\frac {\sqrt 5} 2$. The statement follows immediately.

enter image description here

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    $\begingroup$ Beat me to it as I was making the picture! +1. $\endgroup$ – Deusovi Aug 16 at 17:55
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    $\begingroup$ Could you expand on the assertion in the second sentence of your second paragraph? I was able to confirm it by subtracting the legs of the smallest triangle out of the large diagonals. Is there a more direct way of showing those 2 segments are similar? $\endgroup$ – Simon Aug 17 at 19:11
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    $\begingroup$ @Simon Not sure I understand your doubts. The area of the triangle is 1/4 because there are four of them tiling the full square. Taking the diagonal as the base the hight of each of these triangles must therefore be 2 x 1/4 / length of diagonal. I suspect you are overthinking it. But feel free to prove me wrong ;-) $\endgroup$ – Paul Panzer Aug 17 at 19:49
  • $\begingroup$ That's not the part that seemed like a leap. Sorry if I was unclear--I was trying to avoid spoilers in the comments. It wasn't immediately clear to me why the distance between the parallel diagonals is equal to the height over the diagonal. It took me several steps of reasoning about lengths of segments of the diagonals to reach that conclusion, so I thought your answer could be improved by expanding on that assertion. Or should the similarity of those segments be obvious for a reason I'm missing? $\endgroup$ – Simon Aug 18 at 19:02
  • $\begingroup$ Oh, I see the "easy" way to get there now. Considering the colored triangles in Deusovi's picture, the line intersecting each bisects the diagonal (by definition). That forms a half-scale similar triangle. Therefore the 2 segments in question are congruent. $\endgroup$ – Simon Aug 18 at 21:42
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here is a solution I think using it similarly we can have any desired square fraction.

imgur is slow here

(imp the long grey line is 1st grey line the relatively shorter one is 2nd grey line.)

1. what we do is get the blue lines first by folding in halves multiple time in this case we get 1/8th division.
2. Take five continuous such division from right edge.
3. fold paper to meet the top right corner of the full square and the point which is the bottom end of the 5th blue line (in the image one blue line overlaps the black which is the 4th blue line ).
4. we get the grey line by joining the "end of 5th blue line" and "one corner ". 5. no we have one triangle with sides x and (5/8)*x;
6. Do a similar operation for the second grey line of triangle(with sides x and (3/8)*x) , this time use the endpoint of the 3rd blue line.
7. fold the top edge of paper to get the x/8 length green line which intersects the first grey line and the right edge of the paper.(could be done easily)
8. the region of green line between the 2 grey line is length x/20. >! 9. fold the right edge to get the red line which passes from the point of intersection of the green line and 2nd grey line.
10. now we have this x/20 length measurement on one side which we can copy 4 times by folding the paper to get x/5 length and then make a square.

Now when we have x/5 lenght we will take x/5 length on one edge lets say right edge and 2x/5 length on top edge(thus these 2 length are perpendicular to each other)

this x/sqrt(5) can be used to create a square of area 1/5 of the larger;

imgur is still slow PS: I made a big mistake earlier and got 1/5 th length the edit now give 1/sqrt(5) length

PS: We can generalize it to get any fraction of area if the fraction can be written as sum of 2 sqaures mean here 5 = 22+11 , also if you are really really hardworking you can actually get any desired fractions , but you have to do these last steps multiple time .

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    $\begingroup$ The question asks for a square with 1/5th the area of the original square, not 1/5th the side length. By the way, there is a quicker way to get 1/5th the side length with just two folds: Fold in half side-to-side to create a crease down the middle, making two rectangles. Then fold one rectangle in half along its diagonal, and the corner point marks one fifth of the way across from the opposite side of the square. $\endgroup$ – Jaap Scherphuis Aug 17 at 5:14
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    $\begingroup$ I see I have made a blunder $\endgroup$ – Aakash Mathur Aug 17 at 5:33
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    $\begingroup$ I have done a edit to correct is $\endgroup$ – Aakash Mathur Aug 17 at 6:01
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    $\begingroup$ although a tough process but now we can have any fraction desired , @JaapScherphuis thankyou for pointing out $\endgroup$ – Aakash Mathur Aug 17 at 6:02
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Not an answer. Here is only an animation to visualize the Deusovi nice answer. I hope you enjoy it.

enter image description here

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Extending on Deusovi's answer, you can fold a square to any fraction square of the fraction $n^2/(a^2+b^2)$, where $n <= a-b$.

To achieve $1/5$, choose $n=1$, $a=2$, $b=1$.

Split the edges in $a$ equal parts. Then fold lines "knight-moves" $(a,b)$. This will generate $(a-b)^2$ sqaures of size $1/(a^2+b^2)$. Now gather $n^2$ of these to generate the desired fraction.

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    $\begingroup$ An interesting idea, but the generalised "Split the edges in a equal parts." step isn't trivial when you have no pre-existing marks on the paper and no other tools (unless a is a power of 2) $\endgroup$ – Steve Aug 21 at 9:35
  • $\begingroup$ That's true Steve. I intentionally left that sub-problem out of my answer, to focus on the most interesting part $\endgroup$ – XPlatformer Aug 21 at 10:21
  • $\begingroup$ It wasn't clear what within the "most interesting part" wasn't already adequately covered by the other answers. Perhaps worth highlighting what the new contribution you are making is? $\endgroup$ – Steve Aug 21 at 13:03

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